Question

The specifications for a certain kind of ribbon call for a mean breaking strength of 180 pounds. If five pieces of the ribbon (randomly selected from different rolls) have a mean breaking strength of 169.5 pounds with a standard deviation of 5.7 pounds, test the null hypothesis μ = 180 pounds against the alternative hypothesis μ < 180 pounds at the 0.01 level of significance. Assume that the population distribution is normal.

a) Find the p value

b) Test the hypothesis

c) Find the 99% C

Answer 1

mu = 180 mu < 180 X bar = 169.5, S = 5.7, n = 5 Significance level: alpha = 0.01 Test statistic: T = X bar - mu_0/s/squareroot n = 169.5 - 180/5.7/squareroot 5 = -4.12 the degree of freedom = n - 1 = 4 a) P value for the one-tailed test = 0.0073 b) We can reject the null hypothesis as the P value is less than the significance level of the test. Hence we have evidence that mean breaking strength is less than 180 pounds. c) For 99% confidence and df = 4 T_alpha/2 = 4.6041 CI = X bar plusminus T_alpha/2(S/squareroot n) = 169.5 plusminus 4.6041(5.7/squareroot 5) CI = 169.5 plusminus 11.74 = (157.76, 181.24)

$\mu <180$

$\overline{X}=169.5,S=5.7,n=5$

Significance level: $\alpha =0.01$

Test statistic: $T=\frac{\overline{X}-\mu _0}{\frac{S}{\sqrt{n}}}=\frac{169.5-180}{\frac{5.7}{\sqrt{5}}}=-4.12$

the degree of freedom = n-1=4

a) P value for the one-tailed test =0.0073

b) We can reject the null hypothesis as the P value is less than the significance level of the test.

Hence we have evidence that mean breaking strength is less than 180 pounds.

c)

For 99% confidence and df = 4

$T_{\frac{\alpha }{2}}=4.6041$

$CI=\overline{X}\pm T_{\frac{\alpha }{2}}(\frac{S}{\sqrt{n}})=169.5\pm 4.6041(\frac{5.7}{\sqrt{5}})$

$CI=169.5\pm 11.74=(157.76,181.24)$